Let f(x)=x

We know $[x]$ discontinuous for $x \in Z$

$f(x)=x\left[\frac{x}{2}\right]$ may be discontinuous where $\frac{x}{2}$ is an

integer.

So, points of discontinuity are,

$x=\pm 2, \pm 4, \pm 6, \pm 8$ and 0

but at $x=0$

$\lim _{x \rightarrow 0^{+}} f(x)=0=f(0)=\lim _{x \rightarrow 0^{-}} f(x)$

So, $f(x)$ will be discontinuous at $x=\pm 2, \pm 4, \pm 6$ and $\pm 8$.